Question
The sequence $\left\{ a_{n} \right\},n \in N$ satisfies $a_{1} = 1$ and $5^{a_{n + 1} - a_{n}} = 1 + \frac{1}{n + \frac{2}{3}}$. Then
Options
$\left\lbrack a_{625} \right\rbrack = 4$
(where $\lbrack \cdot \rbrack$ denotes greatest integer function)
Solution
Sol. [Key Concept: Iterate the recurrence; the greatest-integer values follow from the closed form.]
$$\begin{matrix} & \frac{5^{a_{n + 1}}}{5^{a_{n}}} = \frac{3n + 5}{3n + 2} \\ & \frac{5^{a_{2}}}{5^{a_{1}}} \times \frac{5^{a_{3}}}{5^{a_{2}}} \times \ldots\ldots. \times \frac{5^{a_{n}}}{5^{a_{n - 1}}} = \frac{8}{5} \times \frac{11}{8} \times \frac{14}{11} \times \ldots\ldots\frac{3n + 2}{3n - 1} \\ & \frac{5^{a_{n}}}{5^{a_{1}}} = \frac{3n + 2}{5} \Rightarrow 5^{a_{n}} = 3n + 2 \Rightarrow a_{n} = \log_{5}(3n + 2) \\ & \left\lbrack a_{n} \right\rbrack = 3\ \Rightarrow \log_{5}(3n + 2) \in \lbrack 3,4) \\ & \ \Rightarrow n \in \{ 41,42,\ldots\ldots,\ldots\ldots.207\} \\ & \left\lbrack a_{n} \right\rbrack = 4 \Rightarrow \frac{623}{3} \leq n < \frac{3123}{3} \\ & \ \Rightarrow n \in \{ 208,209,\ldots.,1041\} \end{matrix}$$
✘ (a) $\lbrack a_{501}\rbrack = \lbrack\log_{5\ 1505}\rbrack = 4 \neq 3$ (printed key has this wrong)
✔ (b) $\lbrack a_{207}\rbrack = 3$
✔ (c) $\lbrack a_{223}\rbrack = 4$
✔ (d) $\lbrack a_{625}\rbrack = 4$
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